Could anyone point me a program so i can calculate the roots of
$$ K_{ia}(2 \pi)=0 $$
here $ K_{ia}(x) $ is the modified Bessel function of second kind with (pure complex)index 'k' :D
My conjecture of exponential potential means that the solutions are $ s=2a $ with
$$ \zeta (1/2+is)=0. $$
On
I can quantify somewhat Raymond's suggestion that the zeros of $K_{ia}(2\pi)$ are much more regular than the zeros of $\zeta(1/2+i2a)$. The calculations below are rough and I didn't verify the details, so this is perhaps more of a comment than an answer.
The Bessel function in question has the integral representation
$$ K_{ia}(2\pi) = \int_0^\infty e^{-2\pi \cosh t} \cos(at)\,dt. $$
Naively applying the saddle point method to this integral indicates that, for large $a$,
$$ \begin{align} K_{ia}(2\pi) &\approx \frac{e^{-a\pi/2} \sqrt{\pi}}{(a^2-4\pi^2)^{1/4}} \Bigl( \sin f(a) + \cos f(a) \Bigr) \\ &=\frac{e^{-a\pi/2} \sqrt{2\pi}}{(a^2-4\pi^2)^{1/4}} \cos\left( f(a) - \frac{\pi}{4} \right), \end{align} $$
where
$$ f(a) = \frac{a}{2} \log\left(\frac{a^2-2\pi^2+a\sqrt{a^2-4\pi^2}}{2\pi^2}\right) - \sqrt{a^2-4\pi^2}. $$
Here's a plot of $K_{ia}(2\pi)$ in blue versus this approximation in red (both scaled by a factor of $e^{a\pi/2}\sqrt{a}$ as in Raymond's graph).
The approximation has zeros whenever
$$ f(a) = \frac{3\pi}{4} + n\pi, \tag{$*$} $$
and since
$$ f(a) \approx a\log(a/\pi) - a $$
for large $a$ we expect that solutions of $(*)$ for large $n$ satisfy
$$ a \approx \frac{3\pi/4+n\pi}{W\Bigl((3\pi/4+n\pi)/e\Bigr)}, $$
where $W$ is the Lambert W function.
Here's a plot of $e^{a\pi/2} \sqrt{a} K_{ia}(2\pi)$ in blue with these approximate zeros in red.
I used the free pari/gp $(*)$ to get the smallest positive solutions : \begin{array} {l} 9.76877008350997786701461088004548816694073504626\\ 12.4484878927757792253076746584484174280206806290\\ 14.6849597759503320940392747073070219296666773881\\ 16.6915789382924172031614835847470262244558902255\\ 18.5493751970112696445359821473043722645995430537\\ 20.2999030639074716000431186500336768257041209063\\ 21.9679863621154849828379155738628510793020013525\\ 23.5699217186674186162756939876181463960995601839\\ 25.1171217228471128822167760437525949613439934666\\ 26.6179647161774511589115811996122442026801440680\\ 28.0788271377389792376836682977823877359218232495\\ 29.5047018750550818302335792765863272958843396063\\ \end{array}
These solutions are much more regular than the zeta zeros as you may appreciate in this plot of $\;\displaystyle x\mapsto \operatorname{K}_{ix}(2\pi)\;e^{\pi\,x/2}\sqrt{x}\;$ (the oscillations became quickly very small thus the 'rescaling') :
Concerning the relations between the $\zeta\left(\frac 12+ix\right)$ zeros and the $\operatorname{K}_{ix}(2\pi)$ zeros a quick search gave the paper by Fredrik Strömberg "On the zeros of linear combinations of K-Bessel functions" with the following theorem :
The functions \begin{align} F_1(t)&=4\pi^2\left[\operatorname{K}_{\frac 94+\frac{it}2}(2\pi)+\operatorname{K}_{\frac 94-\frac{it}2}(2\pi)\right]\\ F_2(t)&=F_1(t)-6\pi\left[\operatorname{K}_{\frac 54+\frac{it}2}(2\pi)+\operatorname{K}_{\frac 54-\frac{it}2}(2\pi)\right]\\ \end{align} have a total number of zeros in $\{0<\Re(t)\le T\}$ equal to $$N(T)=\frac{T}{2\pi}\log\left(\frac T{2\pi e}\right)+O(\log T),$$ exactly as for $\xi\left(\frac 12+ix\right)$ (and thus $\zeta\left(\frac 12+ix\right)$). Furthermore all of their zeros are located along the real axis.
The function $\xi$ is defined by $\;\displaystyle \xi(s):=\frac 12s(s-1)\pi^{-\frac s2}\Gamma\left(\frac s2\right)\zeta\left(s\right)$
and has the same complex zeros as $\zeta$ but a simpler functional equation : $\xi(s)=\xi(1-s)$.
In fact Hejhal (following Polya's $1926$ work "Bemerkung über die Integraldarstellung der Riemannschen $\xi$-Funktion") studied the approximant : \begin{align} \xi^*_N\left(\frac 12+it\right)=&\sum_{n=1}^N 4\pi^2\,n^4\left[\operatorname{K}_{\frac 94+\frac{it}2}(2\pi n^2)+\operatorname{K}_{\frac 94-\frac{it}2}(2\pi n^2)\right]\\-&\sum_{n=1}^N 6\pi\,n^2\left[\operatorname{K}_{\frac 54+\frac{it}2}(2\pi n^2)+\operatorname{K}_{\frac 54-\frac{it}2}(2\pi n^2)\right]\\ \end{align}
Strömberg adds "Since the approximations are at best loose, one does not seriously propose this as a means of reaching the RH". I don't include the curve(s) obtained because the zeros are rather different (a similar repartition doesn't require the same zeros...) (of course I could too have made a stupid mistake!).
These results and references should be of interest (if you don't already know them) as well as this paper from George Gasper and this and this one by David Cardon. $$-$$ $(*)$ pari/gp's script for the first root :